# How is the trace of the adjoint of Lie algebra elements defined? [duplicate]

Consider an element $$X\in\mathfrak g$$ of some Lie algebra $$\mathfrak g$$. I understand that $$\mathfrak g$$ can be represented via its action on other elements of the same algebra, as $$\operatorname{ad}(X)Y\equiv[X,Y]$$, so that $$\operatorname{ad}(X)\in\operatorname{GL}(\mathfrak g)$$.

One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $$\operatorname{ad}(X)$$.

However, I usually don't see any direct mention of the inner product with respect to which these things are defined. To properly define what something such as $$\operatorname{ad}(X)_{ij}$$ is, don't I need to be able to define uniquely the coefficient of the generator $$X_i$$ in the decomposition of $$[X,X_j]$$ (denoting with $$\{X_i\}$$ the elements of some basis for the algebra)?

Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?

• Trace of a matrix is independent of the basis. – Moishe Kohan Mar 29 '19 at 17:15
• One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384. – Torsten Schoeneberg Mar 29 '19 at 17:16
• @MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace – glS Mar 29 '19 at 17:17
• Surely $\mathfrak{g}$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it? – Torsten Schoeneberg Mar 29 '19 at 17:27
• I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^{-1})=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/… – Moishe Kohan Mar 29 '19 at 18:18

Consider $$\mathfrak{sl}_2$$ with basis $$\{e,f,h\}$$. You have $$ad(e)(e)=0,\;\;\;ad(e)(f)=h,\;\;\;ad(e)(h)=-2e$$ so the matrix of $$ad(e)$$ in this basis is $$\begin{pmatrix}0&0&-2\\0&0&0\\0&1&0\end{pmatrix}$$ which has trace 0.
You can check that the matrix for $$ad(h)$$ is $$\begin{pmatrix}2&0&0\\0&-2&0\\0&0&0\end{pmatrix}$$ which again has trace 0. Can you do $$ad(f)$$?